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3 years ago

A 1.5m wire carries a 6 A current when a potential difference of 68 V is applied. What is the resistance of the wire?

Physics

1 answer:

ANTONII [103]3 years ago

6 0

Answer:

$11.3 \Omega$

Explanation:

We can find the resistance of the wire by using Ohm's law:

$V=RI$

where

V is the voltage applied

R is the resistance

I is the current

In this problem, we know I = 6 A and V = 68 V, so we can re-arrange the equation to find the resistance of the wire:

$R=\frac{V}{I}=\frac{68 V}{6 A}=11.3 \Omega$

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Whenever you push an object across a frictional surface, it starts out taking a lot of force to push to get the object moving. H

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Answer:

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Explanation:

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3 years ago

The tundra is located _______.

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2 years ago

Read 2 more answers

1. A Zambeef delivery track travels 18 km north, 10 km east, and 16 km south. What is its final displacement from the origin?

Amiraneli [1.4K]

Answer:

10.2 km

Explanation:

Displacement is the shortest distance from your original position after movement.

Now, the Zambeef delivery truck moves 18 km north and then 10 km east and then 16 km south.

This means that vertically the truck is 2km short of the 18 km he travelled north..

So we can find the displacement using pythagoras theorem..

Let d be the distance from the current position to the initial position.

Thus;

d² = 2² + 10²

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I have attached a diagram showing the direction of this displacement

3 0

2 years ago

A car is driving 25.5 m/s when it hits

IceJOKER [234]

Answer:

it takes the car 4.362 seconds to cover the distance of 88.4 m.

Explanation:

The distance the car covers is given by the function

$x(t) = vt-\dfrac{1}{2}at^2$ ,

where $v = 25.5m/s$ , and $a = 2.40m/s$ , putting these in we get:

$x(t) = 25.5t-\dfrac{1}{2}(2.4)t^2\\\\x(t) = 25.5t-1.2t^2$

Now, when the car has moved to 88.4m, $x(t) = 88.4$ , or

$x(t) = 25.5t-1.2t^2 = 88.4$

which is a quadratic equation with solutions

$t = 4.362s,\\t = 16.887s$

We take the first solution $t = 4.362s$ , <em>since at that time the car is still moving right and decelerating</em>. The second solution $t =16.887$ describes the situation where the car has stopped decelerating and is now moving leftwards because the decelerating is leftwards, <em>which is utterly wrong because we know that cars do not start moving backwards after the brakes have stopped them! </em>

Thus, it takes the car 4.362 seconds to cover the distance of 88.4 m.

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3 years ago